JEE Main 2022 · 26 Jul · Shift 2 · Physics Q47

← JEE Main 2022 · 26 Jul · Shift 2

PhysicsSingle correct+4 / -1Question 47 of 90

A velocity selector consists of electric field $\vec E=E\hat k$ and magnetic field $\vec B=B\hat j$ with B = 12 mT. The value of E required for an electron of energy 728 eV moving along the positive $x$ -axis to pass undeflected is :

(Given, mass of electron = $9.1\times10^{-31}$ kg)

192 kVm

^{-1}

192 mVm

^{-1}

9600 kVm

^{-1}

16 kVm

^{-1}

Answer and worked solution

Official answer: A

Speed. $v=\sqrt{\frac{2\times728\times1.6\times10^{-19}}{9.1\times10^{-31}}}=1.6\times10^7$ m s $^{-1}$ .

Undeflected when $E=vB=1.6\times10^7\times0.012=1.92\times10^5$ V m $^{-1}$ $=192$ kV m $^{-1}$ .

(A)

Topics:Motion in magnetic and electric fields

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JEE Main 2022 · 26 Jul · Shift 2, Physics question 47